Problem 10 - Precalculus

Question

Compute each of the following. (a) \((2+7 i)(2-7 i)\) (b) \(\frac{-1+3 i}{2+7 i}\) (c) \(\frac{1}{2+7 i}\) (d) \(\frac{1}{2+7 i} \cdot(-1+3 i)\)

Step-by-Step Solution

Answer

(a) 53, (b) \(\frac{19}{53} + \frac{13}{53}i\), (c) \(\frac{2}{53} - \frac{7}{53}i\), (d) \(\frac{19}{53} + \frac{13}{53}i\)."

Step 1: Multiply Complex Conjugates (Part a)

To multiply \((2+7i)\) and \((2-7i)\), we use the formula for products of complex conjugates: \(a^2 + b^2\). Here, \(a = 2\) and \(b = 7\). This results in \(4 + 49 = 53\). Thus, \((2+7i)(2-7i) = 53\).

Step 2: Rationalize the Denominator (Part b)

To compute \(\frac{-1+3i}{2+7i}\), we multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of \(2+7i\) is \(2-7i\). Multiply: \((2+7i)(2-7i) = 53\). Then, multiply the numerator: \((-1+3i)(2-7i) = -1(2) + (-1)(-7i) + (3i)(2) + (3i)(-7i) = -2 + 7i + 6i - 21(-1) = 19 + 13i\). Therefore, \(\frac{-1+3i}{2+7i} = \frac{19+13i}{53}\).

Step 3: Rationalize the Denominator for a Single Fraction (Part c)

For \(\frac{1}{2+7i}\), multiply by the conjugate \(2-7i\). In the denominator, \((2+7i)(2-7i) = 53\). In the numerator: \(1(2-7i) = 2-7i\). Therefore, \(\frac{1}{2+7i} = \frac{2-7i}{53}\).

Step 4: Multiply Simplified Fractions (Part d)

Use the result from Step 3. Multiply \(\frac{1}{2+7i} = \frac{2-7i}{53}\) by \((-1+3i)\). After simplifying: \((2-7i)(-1+3i) = 2(-1) + 2(3i) + (-7i)(-1) + (-7i)(3i) = -2 + 6i + 7i - 21(-1) = 19 + 13i\).Thus \(\frac{1}{2+7i} \cdot (-1+3i) = \frac{19 + 13i}{53} = \frac{19}{53} + \frac{13}{53}i\).

Key Concepts

Multiplying Complex Conjugates

When you multiply two complex conjugates, the process is simpler than it might seem. The conjugate of a complex number reverses the sign of its imaginary part. For example, the conjugate of \(a + bi\) is \(a - bi\). When you multiply these conjugates together, a beautiful cancellation occurs.
Using the special formula for products of conjugates, \((a + bi)(a - bi) = a^2 + b^2\). Notice there's no imaginary part left, just two squared real numbers added together. This means the product is always a real number.
In our exercise, multiplying \((2+7i)\) and \((2-7i)\) involves squaring 2 and 7, giving \(2^2 + 7^2 = 4 + 49 = 53\). This process ensures you eliminate the imaginary parts automatically, streamlining your calculation into a neat and real number result.

Rationalize Denominator

Rationalizing the denominator in a complex fraction means eliminating the imaginary part in the denominator, which involves multiplying both the numerator and denominator by the conjugate of the denominator. This techniques ensures that the fraction is easier to work with.
Let's say you have a fraction like \(\frac{-1+3i}{2+7i}\). You multiply by \(\frac{2-7i}{2-7i}\), which is essentially multiplying the fraction by 1. This change doesn't affect the value, but it makes the expression simpler.
Look at the denominator now: \((2+7i)(2-7i) = 53\). It becomes a real number, removing the imaginary part. In the numerator, you'll use the distributive property:

• \((-1)(2) + (-1)(-7i) + (3i)(2) + (3i)(-7i)\)
• Combine real and imaginary parts separately
• Resulting in \(19 + 13i\)

Thus, the complex fraction \(\frac{-1+3i}{2+7i}\) simplifies to \(\frac{19 + 13i}{53}\), with a rational denominator.

Simplifying Complex Fractions

Simplifying complex fractions often involves applying multiple operations, similar to the example provided in simplifying \(\frac{1}{2+7i}\). You multiply both the numerator and the denominator by the conjugate of the denominator to rid it of its imaginary part.
Performing the multiplication in the denominator, you use \((2+7i)(2-7i) = 53\). It's important because it consolidates the denominator into a real number. For the numerator, simple multiplication of complex numbers results in something like:

• \(1 \cdot (2-7i) = 2 - 7i\)

After multiplication, the fraction is expressed as \(\frac{2 - 7i}{53}\). It appears in a nicely simplified form with the denominator rationalized.
Overall simplification helps in various calculations, allowing easy addition, subtraction, or further multiplication of complex numbers while maintaining clarity in overall mathematical communication.